Question

calculate the ratio of molecules present in 16 g of methane (CH4) and 16 g of oxygen (O2) [atomic masses :C=12,H=1,O=16]
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Answer 1

Answer:

16 g cH4 = 1 mole of ch4 = 6.022*10²³ molecules

16 g O² = 0.5 mole of O² = 3.022*10²³ molecules

ratio is 2:1

Answer 2

Answer:

The ratio = 2 : 1.

Explanation:

Molar mass of methane (CH4) = 16 g mol-1

Molar mass of oxygen (O2) = 32 g mol-1

16 g of CH4 contains = 6.022×10²³ molecules

Now, 32 g of oxygen contains = 6.022×10²³ molecules

16 g of oxygen = 6.022×10²³ ×16 /32

= 3.011×10²³ molecules

Hence, the ratio of molecules present in 16 g of CH4 and 16 g of O2 = 6.022×10²³ / 3.011×10²³

= 2 : 1